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Rebellos
  • Member for 6 years, 7 months
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Proving that $ (\hat{\beta} - \beta)' (X' X) (\hat{\beta} - \beta)$ is independent with SSE
@guy I've grasped all that, the only thing I cannot see implicitly (because I am not that experienced with these expressions) is how I would write them as functions of the quantities mentioned. The rest would be trivial.
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Proving that $ (\hat{\beta} - \beta)' (X' X) (\hat{\beta} - \beta)$ is independent with SSE
@guy Hi, interesting approach (seems clever as well). How does one rigorously elaborate the given hint though ?
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Proving $\text{Var}{(\hat{y}_h)} = \sigma^2 \left(\frac{1}{n} + \frac{(x_h-\bar{x})^2}{S_{xx}}\right)$
Hello and thanks for your input ! I think that this $$\text{Var}(\hat{y}_{x_0}) = \displaystyle\frac{\sigma^2\sum x_i^2}{n\sum(x_i - \bar{x})^2} + \displaystyle\frac{\sigma^2x_0^2}{\sum(x_i - \bar{x})^2} - \displaystyle\frac{2x_0\sigma^2\bar{x}}{\sum(x_i - \bar{x})^2}$$ needs to be proven as well though.
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Showing that $\sum_{i=1}^n (y_i-\hat{y_i})(\hat{y_i} - \bar{y}) = 0$ for the generalized linear model
Interesting and indeed true. I assume the exercise wants us to work under the assumption of Gaussian distributed errors regarding a fitted generalized model, but that isn't mentioned, so I guess your statement is perfect !
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Showing that $\sum_{i=1}^n (y_i-\hat{y_i})(\hat{y_i} - \bar{y}) = 0$ for the generalized linear model
@a_statistician If $ Y = X\beta + \epsilon$ then $\bar{Y} = \bar{X}\beta + \epsilon$ and $\hat{Y} = X\hat{\beta} + \hat{\epsilon}$ and then it is : $$(Y-\hat Y)'(\hat Y - \bar Y) = (X\beta + \epsilon - X\hat{\beta} + \hat{\epsilon})'(X\hat{\beta} + \hat{\epsilon}- \bar{X}\beta + \epsilon)$$ Now, what ?